Recent content by logscale

The thing is we are still treating x as real, which if we use WKB to approximate the asymtotic eigenstate, you will notice that it just blows to infinity when x goes to infinity.That is what I am trying to say, if the Hamiltonian produces non L2 integrable function, then it is a not a well...

Your Hamiltonian certainly looks like having Airy function as a possible solution.

Hey guys I think I know the solution. All Hamiltonian are subjected to boundary condition. If we allow x to be real, then the eigenfunction to this Hamiltonian will not converge when x goes to infinity (therefore we will have problem to show the Hermiticity of -x^4 in momentum representation...

Thanks for the reply. the thing is I don't see why -x^4 is not hermitian. Since we know that p^2 is hermitian (from free particle problem), all we need to do is to investigate -x^4, which has to be hermitian from the integration you have shown. Maybe I miss anything?

Hallo everyone, I have a question, how can I see that the hamiltonian H=p^2-x^4 is not hermitian, with p the momentum operator and x the position operator.